# Is $f(t)=2t\sin(1/t)-\cos(1/t)$ for $t \neq 0$ and $f(t)=0$ for $t=0$ differentiable?

Is the function $f$ defined as:

$f(t)=2tsin(1/t)-cos(1/t)$ for $t \neq 0$

and

$f(t)=0$ for $t=0$

$t \in R$

differentiable?

In my opinion it's not because taking a sequences we can easily check that:

$\lim_{n \to \infty} f(a_n) \neq 0 =f(0)$ therefore $f$ is not continuous at $0$

and not continuous implies not differentiable

Is this correct? Thank you

I also looked at the function on wolfram alpha and it seems that I'm correct but I'm not so sure.

• It is even not continuous. – daulomb Nov 12 '13 at 19:23

Supposing that there is no typing error. The imit of first term is zero but the second one does not exists and hence their difference does not exists. We use the fact that if $\displaystyle\lim_{t\rightarrow a}f(t)$ exists and if $\displaystyle\lim_{t\rightarrow a}g(t)$ d.n.e, then $\displaystyle\lim_{t\rightarrow a}(f(t)+g(t)$ d.n.e.